306. Additive Number

Additive number is a string whose digits can form additive sequence.

A valid additive sequence should contain at least three numbers. Except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two.

For example:

"112358" is an additive number because the digits can form an additive sequence: 1, 1, 2, 3, 5, 8.
1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8

"199100199" is also an additive number, the additive sequence is: 1, 99, 100, 199.
1 + 99 = 100, 99 + 100 = 199

Note: Numbers in the additive sequence cannot have leading zeros, so sequence 1, 2, 03 or 1, 02, 3 is invalid.

Given a string containing only digits '0'-'9', write a function to determine if it's an additive number.

Follow up: How would you handle overflow for very large input integers?

解题思路比较straightforward：

令第一个数字长度为i，第二个数字长度为j，string总长为n。 由于additive number sequence至少有3个数，第三个数是前两个数之和，因此i的范围为[1, n/2]，而且第三个数的长度应该>=max(i,j)。 这样，我们得到了i，j的范围，然后我们采用两层循环来分别判断是否为additive number sequence。 判断isValid的过程与斐波那契数列类似。

public class Solution {
    public boolean isAdditiveNumber(String num) {
        int len = num.length();
        for(int i = 1; i <= len / 2; i++) {
            for(int j = 1; Math.max(i, j) <= len - i - j; j++) {
                if(isValid(num, i, j)) return true;
            } 
        }
        return false;
    }
    
    private boolean isValid(String num, int i, int j) {
        if(num.charAt(0) == '0' && i > 1) return false;
        if(num.charAt(i) == '0' && j > 1) return false;
        Long n1 = Long.valueOf(num.substring(0, i));
        Long n2 = Long.valueOf(num.substring(i, i+j));
        String sum;
        for(int start = i+j; start < num.length(); start += sum.length()) {
            n2 = n1 + n2;
            n1 = n2 - n1;
            sum = n2.toString();
            if(!num.startsWith(sum, start)) return false;
        }
        return true;
    }
}